$A$ galvanometer coil has a resistance of $15\; \Omega$ and the meter shows full-scale deflection for a current of $4\; mA$. How will you convert the meter into an ammeter of range $0$ to $6\; A$?

(A) Resistance of the galvanometer coil,$G = 15\; \Omega$.
Current for which the galvanometer shows full-scale deflection,$I_{g} = 4\; mA = 4 \times 10^{-3}\; A$.
Range of the ammeter required is $I = 6\; A$.
$A$ shunt resistor of resistance $S$ must be connected in parallel with the galvanometer to convert it into an ammeter. The value of $S$ is given by the formula:
$S = \frac{I_{g} G}{I - I_{g}}$
Substituting the values:
$S = \frac{4 \times 10^{-3} \times 15}{6 - 4 \times 10^{-3}}$
$S = \frac{0.06}{6 - 0.004} = \frac{0.06}{5.996} \approx 0.0100066\; \Omega$.
Rounding to significant figures,$S \approx 0.01\; \Omega = 10\; m\Omega$.
Therefore,a shunt resistor of $10\; m\Omega$ must be connected in parallel with the galvanometer.